Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 2, Problem 38P
You have four guinea pigs for a genetic study. One male and one female are from a strain that is pure-breeding for short brown fur. A second male and female are from a strain that is pure-breeding for long white fur. You are asked to perform two different experiments to test the proposal that short fur is dominant to long fur and that brown is dominant to white. You may use any of the four original pure-breed-ing guinea pigs or any of their offspring in experimental matings. Design two different experiments (crossing dif-ferent animals and using different combinations of pheno-types) to test the dominance relationships of alleles for fur length and color, and make predictions for each cross based on the proposed relationships. Anticipate that the litter size will be 12 for each mating and that female guinea pigs can produce three litters in their lifetime.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
In rabbits grey hair is dominant to white & black eyes are dominant to red. -Cross 2 parents who are both heterozygous for both traits.
1st Question to answer ---- How many of the offspring could have grey hair & black eyes?
2nd Question to answer ---- How many of the offspring could have grey hair & red eyes?
3rd Question to answer ---- How many of the offspring could have white hair & black eyes?
4th Question to answer ---- How many of the offspring could have white hair & red eyes?
You have been tasked by a pharmaceutical company to generate a novel strain of Cannabis sativa for therapeutic purposes. Consider a Cannabis strain with three genes (called gene 1, gene 2 and gene 3) that are on different chromosomes and assort independently. Each gene has two alleles, described below:
The T allele for gene 1 is dominant and associated with the trait of producing large amounts of terpenes, and the t allele is recessive and associated with the trait of producing a low amount of terpenes
The S allele for gene 2 is dominant and associated with the trait of fast growth, and the s allele is associated with the trait of slow growth
The F allele for gene 3 is associated with the trait of producing large flowers, and the f allele is associated with the trait of producing small flowers
Assume that an individual plant that is heterozygous for genes 1 and 2, but homozygous ff for gene 3 self-fertilizes.
What is the probability of getting a plant that produces a large number of…
A geneticist has two true-breeding strains of mice. Each strain is homozygous for an independently discovered dominant mutation that causes the mice to have no fur. One mutant strain is called “hairless”, and the other strain is called “naked”. The geneticist crosses hairless and naked mice with each other and the F1 offspring all have no fur. When the F1 mice are crossed with each other, the offspring consist of 187 mutant mice with without fur and 13 normal mice with fur.
a. Are the “hairless” and “naked” mutations alleles of the same gene? Give a reason for your answer.
b. Give the genotypic and associated phenotypic ratios of the F2 offspring. (which genotypes in the offspring of the F1 x F1 cross produce fur and which genotypes produce no fur).
In another strain of mice, coat colour is controlled by a single gene with multiple alleles in a dominance series where cream (A1) > agouti (A2) > brown (A3) > black (A4).
c. Give the genotypes of two phenotypically…
Chapter 2 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
Ch. 2 -
1. Compare and contrast the following terms:
a....Ch. 2 - For the cross , what is the expected genotype...Ch. 2 - 37. Galactosemia is an autosomal recessive...Ch. 2 - In mice, black coat color is dominant to white...Ch. 2 - Two parents plan to have three children. What is...Ch. 2 - Consider the cross AaBbCCAABbCc. a. How many...Ch. 2 - If a chi-square test produces a chi-square value...Ch. 2 -
8. Determine whether the statements below are...Ch. 2 - In the datura plant, purple flower color is...Ch. 2 - 10. The dorsal pigment pattern of frogs can be...
Ch. 2 - 11. Black skin color is dominant to pink skin...Ch. 2 - A male mouse with brown fur color is mated to two...Ch. 2 - 13. Figure 2.12 shows the results of Mendel’s...Ch. 2 - 14. An experienced goldfish breeder receives two...Ch. 2 -
15. The accompanying pedigree shows the...Ch. 2 -
16. A geneticist crosses a pure-breeding strain...Ch. 2 - Suppose an F1 plant from Problem 16 is crossed to...Ch. 2 - 18. In pea plants, the appearance of flowers along...Ch. 2 - 19. If two six-sided dice are rolled, what is the...Ch. 2 - Experimental Insight 2.1 describes data, collected...Ch. 2 -
21. The accompanying pedigree shows the...Ch. 2 - 22. The seeds in bush bean pods are each the...Ch. 2 - List all the different gametes that are possible...Ch. 2 - Organisms with the genotypes AABbCcDd and AaBbCcDd...Ch. 2 -
52. In humans, the ability to bend the thumb...Ch. 2 - In the fruit fly Drosophila, a rudimentary wing...Ch. 2 - In pea plants, plant height, seed shape, and seed...Ch. 2 - A variety of pea plant called Blue Persian...Ch. 2 - 29. In tomato plants, the production of red fruit...Ch. 2 - A male and a female are each heterozygous for both...Ch. 2 - 31. In a sample of families with children each,...Ch. 2 - Prob. 32PCh. 2 - A woman expressing a dominant phenotype is...Ch. 2 - Two parents who are each known to be carriers of...Ch. 2 - 33. An organism having the genotype AaBbCcDdEe is...Ch. 2 - 34. A man and a woman are each heterozygous...Ch. 2 - For a single dice roll, there is a 16 chance that...Ch. 2 - You have four guinea pigs for a genetic study. One...Ch. 2 - 37. Galactosemia is an autosomal recessive...Ch. 2 - Sweet yellow tomatoes with a pear shape bring a...Ch. 2 - A cross between a spicy variety of Capsicum annum...Ch. 2 - Alkaptonuria is an infrequent autosomal recessive...Ch. 2 - 41. Humans vary in many ways from one another....Ch. 2 - 42. In chickens, the presence of feathers on the...Ch. 2 -
43. A pure-breeding fruit fly with the...Ch. 2 - 44. Situs inversus is a congenital condition in...Ch. 2 - 45. Domestic dogs evolved from ancestral grey...Ch. 2 - Alleles of the IGF-1 gene in dogs, encoding...Ch. 2 - 49. The Basalt Seed Lending Library run by the...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Tomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Convert the expected phenotypic ratio from Part A into the expected probability for each of the four phenotypes and record them in the table below. Calculate the probability for each of the four phenotypes observed in the cross from the data presented at the beginning of the question by dividing the number of progeny in each class by the total number of progeny and record these in the table below. Compare the expected probabilities of each phenotype to the observed probabilities. Are the gene for fruit colour and…arrow_forwardTomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Convert the expected phenotypic ratio from Part A into the expected probability for each of the four phenotypes and record them in the table below. Calculate the probability for each of the four phenotypes observed in the cross from the data presented at the beginning of the question by dividing the number of progeny in each class by the total number of progeny and record these in the table below. Probability = Number of Progeny in Phenotype Class ÷ Total Progenyarrow_forwardTomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Convert the expected phenotypic ratio from Part A into the expected probability for each of the four phenotypes and record them in the table below. Calculate the probability for each of the four phenotypes observed in the cross from the data presented at the beginning of the question by dividing the number of progeny in each class by the total number of progeny and record these in the table below. Probability = Number of Progeny in Phenotype Class ÷ Total Progeny Compare the expected probabilities of each phenotype to…arrow_forward
- Tomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Part A: Record the phenotypes and the phenotype ratio in lowest terms. Part B: Convert the expected phenotypic ratio from Part A into the expected probability for each of the four phenotypes and record them in the table below. Calculate the probability for each of the four phenotypes observed in the cross from the data presented at the beginning of the question by dividing the number of progeny in each class by the total number of progeny and record these in the table below. (Probability = Number of Progeny in Phenotype…arrow_forwardTomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Part A: Record the phenotypes and the phenotype ratio in lowest terms. Part B: Convert the expected phenotypic ratio from Part A into the expected probability for each of the four phenotypes and record them in the table below. Calculate the probability for each of the four phenotypes observed in the cross from the data presented at the beginning of the question by dividing the number of progeny in each class by the total number of progeny and record these in the table below. Probability = Number of Progeny in Phenotype…arrow_forwardHemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?arrow_forward
- Pedigree Analysis Is a Basic Method in Human Genetics Using the pedigree provided, answer the following questions. a. Is the proband male or female? b. Is the grandfather of the proband affected? c. How many siblings does the proband have, and where is he or she in the birth order?arrow_forwardPedigree analysis is a fundamental tool for investigating whether or not a trait is following a Mendelian pattern of inheritance. It can also be used to help identify individuals within a family who may be at risk for the trait. Adam and Sarah, a young couple of Eastern European Jewish ancestry, went to a genetic counselor because they were planning a family and wanted to know what their chances were for having a child with a genetic condition. The genetic counselor took a detailed family history from both of them and discovered several traits in their respective families. Sarahs maternal family history is suggestive of an autosomal dominant pattern of cancer predisposition to breast and ovarian cancer because of the young ages at which her mother and grandmother were diagnosed with their cancers. If a mutant allele that predisposed to breast and ovarian cancer was inherited in Sarahs family, she, her sister, and any of her own future children could be at risk for inheriting this mutation. The counselor told her that genetic testing is available that may help determine if this mutant allele is present in her family members. Adams paternal family history has a very strong pattern of early onset heart disease. An autosomal dominant condition known as familial hypercholesterolemia may be responsible for the large number of deaths from heart disease. As with hereditary breast and ovarian cancer, genetic testing is available to see if Adam carries the mutant allele. Testing will give the couple more information about the chances that their children could inherit this mutation. Adam had a first cousin who died from Tay-Sachs disease (TSD), a fatal autosomal recessive condition most commonly found in people of Eastern European Jewish descent. Because TSD is a recessively inherited disorder, both of his cousins parents must have been heterozygous carriers of the mutant allele. If that is the case, Adams father could be a carrier as well. If Adams father carries the mutant TSD allele, it is possible that Adam inherited this mutation. Because Sarah is also of Eastern European Jewish ancestry, she could also be a carrier of the gene, even though no one in her family has been affected with TSD. If Adam and Sarah are both carriers, each of their children would have a 25% chance of being afflicted with TSD. A simple blood test performed on both Sarah and Adam could determine whether they are carriers of this mutation. Would you decide to have a child if the test results said that you carry the mutation for breast and ovarian cancer? The heart disease mutation? The TSD mutation? The heart disease and the mutant alleles?arrow_forwardPedigree analysis is a fundamental tool for investigating whether or not a trait is following a Mendelian pattern of inheritance. It can also be used to help identify individuals within a family who may be at risk for the trait. Adam and Sarah, a young couple of Eastern European Jewish ancestry, went to a genetic counselor because they were planning a family and wanted to know what their chances were for having a child with a genetic condition. The genetic counselor took a detailed family history from both of them and discovered several traits in their respective families. Sarahs maternal family history is suggestive of an autosomal dominant pattern of cancer predisposition to breast and ovarian cancer because of the young ages at which her mother and grandmother were diagnosed with their cancers. If a mutant allele that predisposed to breast and ovarian cancer was inherited in Sarahs family, she, her sister, and any of her own future children could be at risk for inheriting this mutation. The counselor told her that genetic testing is available that may help determine if this mutant allele is present in her family members. Adams paternal family history has a very strong pattern of early onset heart disease. An autosomal dominant condition known as familial hypercholesterolemia may be responsible for the large number of deaths from heart disease. As with hereditary breast and ovarian cancer, genetic testing is available to see if Adam carries the mutant allele. Testing will give the couple more information about the chances that their children could inherit this mutation. Adam had a first cousin who died from Tay-Sachs disease (TSD), a fatal autosomal recessive condition most commonly found in people of Eastern European Jewish descent. Because TSD is a recessively inherited disorder, both of his cousins parents must have been heterozygous carriers of the mutant allele. If that is the case, Adams father could be a carrier as well. If Adams father carries the mutant TSD allele, it is possible that Adam inherited this mutation. Because Sarah is also of Eastern European Jewish ancestry, she could also be a carrier of the gene, even though no one in her family has been affected with TSD. If Adam and Sarah are both carriers, each of their children would have a 25% chance of being afflicted with TSD. A simple blood test performed on both Sarah and Adam could determine whether they are carriers of this mutation. Would you want to know the results of the cancer, heart disease, and TSD tests if you were Sarah and Adam? Is it their responsibility as potential parents to gather this type of information before they decide to have a child?arrow_forward
- Pedigree analysis is a fundamental tool for investigating whether or not a trait is following a Mendelian pattern of inheritance. It can also be used to help identify individuals within a family who may be at risk for the trait. Adam and Sarah, a young couple of Eastern European Jewish ancestry, went to a genetic counselor because they were planning a family and wanted to know what their chances were for having a child with a genetic condition. The genetic counselor took a detailed family history from both of them and discovered several traits in their respective families. Sarahs maternal family history is suggestive of an autosomal dominant pattern of cancer predisposition to breast and ovarian cancer because of the young ages at which her mother and grandmother were diagnosed with their cancers. If a mutant allele that predisposed to breast and ovarian cancer was inherited in Sarahs family, she, her sister, and any of her own future children could be at risk for inheriting this mutation. The counselor told her that genetic testing is available that may help determine if this mutant allele is present in her family members. Adams paternal family history has a very strong pattern of early onset heart disease. An autosomal dominant condition known as familial hypercholesterolemia may be responsible for the large number of deaths from heart disease. As with hereditary breast and ovarian cancer, genetic testing is available to see if Adam carries the mutant allele. Testing will give the couple more information about the chances that their children could inherit this mutation. Adam had a first cousin who died from Tay-Sachs disease (TSD), a fatal autosomal recessive condition most commonly found in people of Eastern European Jewish descent. Because TSD is a recessively inherited disorder, both of his cousins parents must have been heterozygous carriers of the mutant allele. If that is the case, Adams father could be a carrier as well. If Adams father carries the mutant TSD allele, it is possible that Adam inherited this mutation. Because Sarah is also of Eastern European Jewish ancestry, she could also be a carrier of the gene, even though no one in her family has been affected with TSD. If Adam and Sarah are both carriers, each of their children would have a 25% chance of being afflicted with TSD. A simple blood test performed on both Sarah and Adam could determine whether they are carriers of this mutation. If Sarah carries the mutant cancer allele and Adam carries the mutant heart disease allele, what is the chance that they would have a child who is free of both diseases? Are these good odds?arrow_forwardTomato Plants In tomato plants, round fruit (R) is dominant to oval fruit (r). Pure breeding plants with red and round fruit (FFRR) were crossed to pure breeding plants with yellow and oval fruit (ffrr). The red and round F1 progeny were then testcrossed to plants that were homozygous recessive for both genes (ffrr) with the following results: Phenotypes Number of Offspring Red and round 2 255 Red and oval 290 Yellow and round 310 Yellow and oval 2 145 Complete a Punnett square to determine the expected testcross progeny on the basis of Mendel’s Law of Independent Assortment. Record the phenotypes and the phenotype ratio in lowest terms.arrow_forwardA recessive maternal effect mutant in zebrafish, called ichabod, results in embryos lacking heads that are non-viable. You have been instructed to identify females that are homozygous for the ichabod mutant allele. At your disposal are a tank of wild-type fish (males and females), a tank of male and female parental fish that are all heterozygous for the ichabod mutant allele (ichabod/+), and a tank of F1 fish derived from a cross between a heterozygous male and heterozygous female (ichabod/+). Which of the following would be a way to identify females that are homozygous mutant, i.e. ichabod/ichabod? Select all answers that would work.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
Human Heredity: Principles and Issues (MindTap Co...
Biology
ISBN:9781305251052
Author:Michael Cummings
Publisher:Cengage Learning
How to solve genetics probability problems; Author: Shomu's Biology;https://www.youtube.com/watch?v=R0yjfb1ooUs;License: Standard YouTube License, CC-BY
Beyond Mendelian Genetics: Complex Patterns of Inheritance; Author: Professor Dave Explains;https://www.youtube.com/watch?v=-EmvmBuK-B8;License: Standard YouTube License, CC-BY